Chapter 2: Problem 54
Suppose \(g\) is the function whose domain is the interval \([1, \infty)\) with $$ g(x)=x^{2}+4 x+7 $$ (a) What is the range of \(g\) ? (b) Find a formula for \(g^{-1}\). (c) What is the domain of \(g^{-1}\) ? (d) What is the range of \(g^{-1}\) ?
Short Answer
Expert verified
(a) The range of \(g(x)\) is [12, ∞).
(b) The inverse function is \(g^{-1}(x) = \frac{-4 \pm \sqrt{4x - 12}}{2}\).
(c) The domain of \(g^{-1}(x)\) is [3, ∞).
(d) The range of \(g^{-1}(x)\) is [1, ∞).
Step by step solution
01
Analyze the function g(x)
Observe that the function \(g(x) = x^2 + 4x + 7\) is a quadratic function. The coefficient of the \(x^2\) term is positive, so the parabola opens upwards.
02
Find the vertex of the parabola
To find the vertex of the parabola, use the formula \(h = \frac{-b}{2a}\), where a and b are the coefficients of the \(x^2\) and x terms, respectively. In our case, \(a = 1\) and \(b = 4\).
Calculate for h:
\[ h = \frac{-4}{2(1)} = -2 \]
Now, substitute h in the function g(x) to find the y-coordinate of the vertex.
\[ k = g(-2) = (-2)^2 + 4(-2) + 7 = 4 - 8 + 7 = 3 \]
The vertex of the parabola is (-2, 3), but notice that x = -2 is not within the domain of g(x).
03
Find the range of g(x) based on the domain
Since the domain of g(x) is [1, ∞) and we know that g(x) is an upward-opening parabola, we can infer that the values of g(x) range from the lowest point on the graph at the given domain x=1 to infinity. Calculate g(1) to find the range of g(x).
\[ g(1) = 1^2 + 4(1) + 7 = 1 + 4 + 7 = 12 \]
Thus, the range of g(x) is [12, ∞).
04
Find the inverse function g^(-1)(x)
To find the inverse function, switch x and y in the function g(x) and solve for y:
\[x = y^2 + 4y +7\]
Subtract x from both sides:
\[y^2 + 4y + 7 - x = 0\]
Now, we use the quadratic formula to solve for y:
\[y= \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
In this case, \(a=1, b=4\), and \(c=7-x\).
Plug in the values and simplify:
\[ y = \frac{-4 \pm \sqrt{4^2 -4(1)(7-x)}}{2(1)} \]
\[ y = \frac{-4 \pm \sqrt{16 - 28 + 4x}}{2} \]
So, the inverse function is \(g^{-1}(x) = \frac{-4 \pm \sqrt{4x - 12}}{2}\).
05
Find the domain of g^(-1)(x)
Observe that the expression inside the square root should be non-negative. Set the expression to greater than or equal to zero and solve for x:
\[4x - 12 \geq 0\]
\[4x \geq 12\]
\[x \geq 3\]
So, the domain of \(g^{-1}(x)\) is [3, ∞).
06
Find the range of g^(-1)(x)
Since the domain of \(g^{-1}(x)\) is [3, ∞), we can find its range by looking at the inverse mapping of g(x). Remember that the range of g(x) is [12, ∞). By the properties of inverse functions, the range of \(g^{-1}(x)\) is the domain of the function g(x):
Thus, the range of \(g^{-1}(x)\) is [1, ∞).
Solution:
(a) The range of \(g(x)\) is [12, ∞).
(b) The inverse function is \(g^{-1}(x) = \frac{-4 \pm \sqrt{4x - 12}}{2}\).
(c) The domain of \(g^{-1}(x)\) is [3, ∞).
(d) The range of \(g^{-1}(x)\) is [1, ∞).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Function
A quadratic function is a type of polynomial function where the highest exponent of the variable is 2. It is generally written in the form:
The focal point of a parabola, known as the vertex, is particularly important. It denotes the minimum or maximum value of the function. This vertex can be found using the formula:
- \( f(x) = ax^2 + bx + c \)
The focal point of a parabola, known as the vertex, is particularly important. It denotes the minimum or maximum value of the function. This vertex can be found using the formula:
- \( h = \frac{-b}{2a} \)
Range of a Function
The range of a function is the set of possible outputs or y-values that a function can produce. For quadratic functions, the range is largely influenced by the vertex and the direction in which the parabola opens.
In the function \( g(x) = x^2 + 4x + 7 \), the parabola opens upwards, as indicated by the positive coefficient of \( x^2 \). Therefore, the range begins from the lowest y-value, occurring at the smallest x-value within the domain, extending to positive infinity.
Given the domain is restricted to \([1, \infty)\), substituting \( x = 1 \) into the function gives the lowest point within the domain, \( g(1) = 12 \). Thus, the range is from 12 to infinity: \([12, \infty)\).
In the function \( g(x) = x^2 + 4x + 7 \), the parabola opens upwards, as indicated by the positive coefficient of \( x^2 \). Therefore, the range begins from the lowest y-value, occurring at the smallest x-value within the domain, extending to positive infinity.
Given the domain is restricted to \([1, \infty)\), substituting \( x = 1 \) into the function gives the lowest point within the domain, \( g(1) = 12 \). Thus, the range is from 12 to infinity: \([12, \infty)\).
Domain and Range
The domain encompasses all possible input values (x-values) for a function, whereas the range covers all potential output values (y-values).
When considering inverses, it's essential to understand that the range of a function becomes the domain of its inverse, and vice versa. For the function \( g(x) = x^2 + 4x + 7 \) with domain \([1, \infty)\) and range \([12, \infty)\), its inverse function \( g^{-1}(x) \) retains this relationship.
When considering inverses, it's essential to understand that the range of a function becomes the domain of its inverse, and vice versa. For the function \( g(x) = x^2 + 4x + 7 \) with domain \([1, \infty)\) and range \([12, \infty)\), its inverse function \( g^{-1}(x) \) retains this relationship.
- The domain of \( g^{-1}(x) \) is \([12, \infty)\)
- The range of \( g^{-1}(x) \) is \([1, \infty)\).
Quadratic Formula
The quadratic formula is a powerful tool used to solve quadratic equations, particularly when the factorization method is not feasible. It is derived from the quadratic function:
In the case of finding an inverse function such as \( g^{-1}(x) \), the quadratic formula assists in expressing \( y \) in terms of \( x \) when given \( x = y^2 + 4y + 7 \). Applying the formula enables us to simplify this to:
- \( ax^2 + bx + c = 0 \).
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
In the case of finding an inverse function such as \( g^{-1}(x) \), the quadratic formula assists in expressing \( y \) in terms of \( x \) when given \( x = y^2 + 4y + 7 \). Applying the formula enables us to simplify this to:
- \( g^{-1}(x) = \frac{-4 \pm \sqrt{4x - 12}}{2} \)